Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1

For $A=(a_{ij})\in M_n(\mathbb{C})$, the determinant isForany nonzero term in the Leibniz formula for determinants must match every row in the lower block to a column in the $N$ block. The remaining upper rows must then match the $M$ columns. The sum consequently factors into the determinant sums for the two diagonal blocks:

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Move the first $n-k$ columns ofpast its last $k$ columns. This requires $k(n-k)$ column interchanges and produces the block-diagonal matrixEach interchange reverses the determinant's sign, so part (a) gives

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A unit-speed curve on a smooth embedded surface is a geodesic exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.

Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the formIndeed, unrolling the cylinder to its universal cover ${\mathbb{R}}^2$ turns these curves into straight lines. Directly,which is parallel to the cylinder's radial normal vector, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.

Yes. In polar coordinates $(r,\theta )$ on ${\mathbb{R}}^2\setminus \{0\}$, use the Riemannian metricThe coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb{R}\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.

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The Taylor series of the complex sine isTherefore, for $z\ne 0$,and the right side defines a holomorphic function at $z=0$ with value $1$. Thus the apparent singularity is a removable singularity.

The resulting power series isThe ratio test shows convergence for every $z\in \mathbb{C}$, so its radius of convergence is

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Forthe Euler-Lagrange equation isHere this becomesBecause the forcing is resonant with the complementary solution, a particular integral is $(x/2)\cos x$. HenceandThe Neumann boundary conditions giveThus

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Let the columns of $A$ be $a_1,a_2,a_3$. The Gram-Schmidt process givesNext,soFor the final column,andThus $q_3=\frac{1}{2}(-1,-1,1,-1{)}^T$ and $r_{33}=2$. The reduced QR decomposition isThe columns of $Q$ are orthonormal, and direct multiplication gives $QR=A$.

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The Rao-Blackwell theorem says that if $U$ is an estimator with finite second moment and $T$ is a sufficient statistic, thenhas the same expectation as $U$ and satisfieswith equality only when $U$ is already a function of $T$ almost surely.

Here $\hat{\theta }=1_{\{X_1=1\}}$ is unbiased becauseConditional on $T=t$, every weak composition $(x_1,\dots ,x_n)$ of $t$ has the same probability $p^n(1-p{)}^t$. There aresuch compositions. For $t\ge 1$, those with $x_1=1$ correspond to weak compositions of $t-1$ into $n-1$ parts, of which there areThe Rao-Blackwell estimator is thereforeFor $n\ge 2$ and $0<p<1$, $\hat{\theta }$ is not determined by $T$; for example, conditional on $T=1$, the sole failure can occur in any coordinate. The variance inequality is consequently strict:

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Starting from $x_0\in {\mathbb{R}}^n$, gradient descent repeatedly computes the gradient and updatesstopping when the gradient norm, step, or objective decrease is sufficiently small.

The Hessian bounds say that $f$ is $\alpha$-strongly convex and has $\beta$-smooth gradient. With $\eta =1/\beta$,Thus the iteration count isconvergence becomes slower linearly with the condition number $\kappa =\beta /\alpha$.

Forthe Hessian matrix is $\mathrm{diag}(2,200,20000)$, soTakeThenwhose Hessian is $2I$ and whose condition number is $1$.

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The adjoint operator ${\alpha }^{\ast }$ is defined bySince $V$ is a nonzero finite-dimensional complex vector space, $\alpha$ has an eigenvector $v$ with eigenvalue $\lambda$. Normality givesfor every $x$, by expanding both sides and using $\alpha {\alpha }^{\ast }={\alpha }^{\ast }\alpha$. Taking $x=v$ yields

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For the common eigenvector $v$, its orthogonal complement $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. Indeed, if $w\perp v$, thenand the analogous calculation applies to ${\alpha }^{\ast }w$. The restriction of $\alpha$ to $v^{\perp }$ is therefore again normal. Induction on $\dim V$, beginning with the normalized vector $v/\Vert v\Vert$, produces an orthonormal eigenbasis. This proves the finite-dimensional spectral theorem for normal operators.

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If $Av=\lambda v$ for a nonzero real vector $v$, then skew-symmetry givesThusso a real skew-symmetric matrix has no nonzero real eigenvalue.

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Over $\mathbb{C}$, the real skew-symmetric matrix $A$ is normal because $A^{\ast }=-A$. Part (a) gives an orthonormal complex eigenbasis, and the nonzero eigenvalues are purely imaginary pairs $\pm i\lambda$.

For each $\lambda >0$, choose a unit real vector $x$ in the kernel of $A^2+{\lambda }^{2}I$ and putThen $x\perp y$, $\Vert y\Vert =1$, andThus $A$ has matrixon the orthonormal basis $(x,y)$. Distinct such invariant planes are orthogonal; complete them by an orthonormal basis of $\ker A$. Taking these basis vectors as the columns of an orthogonal matrix $R$ gives

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A Euclidean domain is an integral domain $R$ equipped with a function $\delta :R\setminus \{0\}\to \mathbb{N}$ such that for $a,b\in R$, $b\ne 0$, there are $q,r\in R$ withFor the Gaussian integers $\mathbb{Z}[i]$, take $\delta (z)=N(z)=|z{|}^2$. Choosing a Gaussian integer nearest to $a/b$ makes the remainder norm smaller than $N(b)$.

The units are precisely the elements of norm one:Unique factorization in this Euclidean domain givesThe displayed factors have prime norms $2$ or $5$, so they are irreducible; factors appearing together are nonassociate.

Now suppose $x^2+4=y^3$. Necessarily $y>0$. First let $y$ be odd. Then $x$ is odd, and $x+2i$ and $x-2i$ are coprime in $\mathbb{Z}[i]$: a common Gaussian prime would divide $4i$, while their product has odd norm. Henceup to a unit, which can be absorbed into the cube. Comparing imaginary parts givesChecking $b\mid 2$ yields only $b=-2$, $a=\pm 1$, and therefore

If $y$ is even, congruence modulo $8$ gives $x=2X$ with $X$ odd and $y=2Y$, whereEach of $X\pm i$ contains exactly one factor $1+i$, so the coprime quotients are cubes up to units. ThusComparing the imaginary part after the four possible units reduces toEach integer factor must have absolute value one. Substitution then gives $a^2+b^2=1$, so $Y=1$ and $X=\pm 1$. HenceAll four pairs satisfy the equation, so the complete answer is

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This is always true. Given $\epsilon >0$, choose $N_A,N_B$ such thaton $A$ for $n\ge N_A$ and on $B$ for $n\ge N_B$. For $n\ge \max (N_A,N_B)$ the same inequality holds at every point of $X=A\cup B$. Hence $f_n\to f$ uniformly on $X$.

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This may be false when the limiting functions are unbounded. On $X=\mathbb{R}$, letBoth sequences converge uniformly to their stated limits, butis unbounded for every $n$. Thus the products do not converge uniformly.

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For every $(x,y)\in [1,2{]}^2$,Moreover,The bound is independent of $(x,y)$ and tends to zero, so

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This is always true. Uniform convergence makes $(f_n)$ a uniformly Cauchy sequence: for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$ and $m,n\ge N$. Letting $x\to x_0$ through points of $A\setminus \{x_0\}$ and using the assumed limits givesThus $(y_n)$ is a Cauchy sequence. Since $Y$ is a complete metric space, $(y_n)$ converges in $Y$.

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This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such thatfor every $x\in X$ and $n\ge N$. The continuous function $g$ is uniformly continuous on the compact interval $[-M-1,M+1]$. Therefore, given $\epsilon >0$, a sufficiently small uniform bound on $|f_n-f|$ impliesHence $g\circ f_n\to g\circ f$ uniformly.

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Every orientation-preserving isometry of the Poincare half-plane model has the formwith matrices identified up to multiplication by $-I$. Thus $G\cong PSL(2,\mathbb{R})$.

For $z=x+iy\in \mathbb{H}$, the affine map $(z-x)/y$ sends $z$ to $i$, proving transitivity on points. Real translations and positive dilations act transitively on finite boundary points, while inversion $z\mapsto -1/z$ exchanges $0$ and $\infty$; hence the action on $\mathbb{R}\cup \{\infty \}$ is transitive. A hyperbolic line is determined by its unordered pair of boundary endpoints, and a real Möbius transformation can send any such pair to $\{0,\infty \}$. Therefore $G$ is also transitive on hyperbolic lines.

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By point transitivity, it suffices to take $P=i$. The conditionis equivalent to $d=a$, $c=-b$, and the determinant condition becomes $a^2+b^2=1$. Thus the stabilizer consists ofmodulo $\pm I$. Writing $a=\cos \phi$, $b=\sin \phi$ identifies it with rotations of the tangent plane, and the doubled matrix angle removes the $\pm I$ quotient. Hence

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An isometry sends the geodesic through $P,Q$ to the imaginary axis and sends the two points to $ir$ and $i/r$ for some $r>0$. The transformationbelongs to $PSL(2,\mathbb{R})$ and exchanges $ir$ with $i/r$. Conjugating it back gives an orientation-preserving isometry exchanging $P$ and $Q$.

For uniqueness, the quotient of two such isometries fixes both $P$ and $Q$. An orientation-preserving hyperbolic isometry fixing two distinct interior points fixes their connecting geodesic and both tangent directions there, hence is the identity. Therefore the exchanging isometry is unique; geometrically it is the hyperbolic half-turn about the midpoint of the segment $PQ$.

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Use an orientation-preserving isometry to send $P$ to $i$ and $\ell$ to the imaginary axis, whose boundary points are $0,\infty$. A geodesic $m$ through $i$ meeting it at angle $\theta$ has boundary pointsIndeed $uv=-1$, so the semicircle with endpoints $u,v$ passes through $i$, and its tangent there makes angle $\theta$ with the vertical.

The cross-ratio is invariant under the normalizing Möbius transformation. With the suitable ordering $(u,v,\infty ,0)$ and the convention used here,

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The hypothesis makes $\Omega$ star-shaped with centre $z_0$. For small $h$ with the triangle having vertices $z_0,w,w+h$ contained in $\Omega$, the vanishing triangular integral givesParametrizing the final segment,by continuity. Thus $g$ is holomorphic andThis is the triangle-integral construction behind Morera's theorem.

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The set $D$ is star-shaped with centre $z_0$: if $z\in D$, every point of $[z_0,z]$ has its segment to $z_0$ contained in the same segment and therefore avoids the origin. Applying part (a) and the stated triangular-integral fact to $1/z$ gives a holomorphic functionChoose either square root $c$ of $z_0$ and defineThen $h$ is holomorphic. Moreover,and its value at $z_0$ is one. Hence

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Because $\cos 0=1$, continuity gives a disc $U$ about zero on which $\cos z\ne 0$. The function $-\tan z=(\log \cos z{)}^{\prime }$ has a holomorphic primitive $G$ on a sufficiently small disc, normalized by $G(0)=0$. Thenis holomorphic and satisfies $f(0)=1$ and $f(z{)}^2=\cos z$ on $U$.

No such square root exists throughout $|z|<2$. The point $z=\pi /2$ lies in that disc and is a simple zero of $\cos z$, since $-\sin (\pi /2)=-1$. Every zero of the square of a holomorphic function has even order of a zero of a holomorphic function, contradicting this simple zero.

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Take positive $y$ vertically downward. For a short string element, the transverse tension resultant is $T{y}_{xx}dx$, while gravity and linear drag contribute $\mu gdx$ and $-2k\mu y_{t}dx$. Newton's second law givesSince $T=\mu c^2$, the damped wave equation isChoosing upward displacement reverses the sign of the gravity term.

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With gravity omitted, the initial data contain only the first fixed-end normal mode, so writeThenBecause $k=\pi c/l$, this oscillator is critically damped, andTherefore

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The first mode remains the solution from part (b). The applied force excites only the third mode, sowhere, using $\pi c/l=k$,A particular solution isAdding the decaying complementary solution and imposing the initial conditions gives

As $t\to \infty$, all transients decay but the driven third-mode oscillation remains. The string approaches a periodic steady-state response, rather than coming to rest.

With gravity omitted, its mechanical energy isWriting $s=\alpha /(34k^2)$ and letting $t\to \infty$ givesThe limiting energy is periodic because the external force continually supplies the energy dissipated by drag.

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For the one-dimensional quantum harmonic oscillator,This is the Time-dependent Schrodinger equation with harmonic potential.

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Fordirect differentiation givesandSubstitution into the Schrodinger equation and equality of the coefficients of $x^2,x,1$ give the necessary and sufficient Riccati equation system

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Put $u=a+i\omega t$. ThenThe $G$ equation from part (b) is satisfied whensoWith this value, the $F$ equation follows from ${\tanh }^{2}u=1-{\mathrm{sech}}^{2}u$: both its constant and ${\mathrm{sech}}^{2}u$ coefficients agree. Thus the stated $F,G$ satisfy both shape equations; the remaining scalar equation determines the normalization factor $N(t)$.

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The probability density isCompleting the square shows that, when $\mathrm{Re}F>0$,where

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Writing $\theta =\omega t$ andthe supplied identities giveUsing $A=m\omega /(2\hslash )$ in part (d),The peak executes simple harmonic motion about the origin with angular frequency $\omega$ and amplitude $\sqrt{\hslash /(m\omega )}/\sinh a$.

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At each fixed time, $|\psi (x,t){|}^2$ is a bell-shaped Gaussian centred at $x=h(t)$, with width determined by $\mathrm{Re}F(t)>0$. Completing the square as in part (d) makes the density symmetric about $h(t)$. Therefore its position expectation value isThe sketch is a Gaussian peak whose centre oscillates between the two turning positions found in part (e), while its width varies periodically through $\mathrm{Re}F(t)$.

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For a coaxial Gaussian cylinder of radius $r$ and length $L$, symmetry makes the electric field radial and constant on the curved surface. Gauss's law givessoSince $\mathbf{E}=-\nabla \varphi$, a potential relative to an arbitrary reference radius $r_0$ isAn infinite line charge has no finite convention $\varphi (\infty )=0$.

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Take the wire at $(a,0)$ to carry $+\lambda$ and that at $(-a,0)$ to carry $-\lambda$. Ifthen, up to one additive constant,Thus an equipotential with $k=2\pi {\epsilon }_0\varphi /\lambda$ satisfies $r_{-}/r_{+}=e^k$. For $k\ne 0$, completing the square gives the Apollonius circleIts centre is $(a\coth k,0)$ and its radius is $a/|\sinh k|$. Positive and negative values give nested circles around the positive and negative wires. The electric field is orthogonal to these circles and points from the positive wire toward the negative wire. For $\varphi =0$, the limiting equipotential is the straight line $x=0$.

Direct superposition givesIn the limit $a\to 0$ with $\lambda a=p$, the potential and field become those of a two-dimensional electric dipole:and